The parametric equations of a line are given below - HSC - SSCE Mathematics Extension 1 - Question 11 - 2023 - Paper 1

To find the number of arrangements of the letters in 'CONDOBOLIN', we note that the word contains 11 letters.

In the word, we have:

• C: 1
• O: 2
• N: 1
• D: 1
• B: 1
• L: 1
• I: 1

The total arrangements can be calculated using the formula for permutations of multiset: rac{n!}{n_1! \cdot n_2! \cdots n_k!} where:

• $n$ = total number of items (letters),
• $n_1$, $n_2$, ..., $n_k$ = counts of each indistinguishable item.

Thus, we have: $\text{Arrangements} = \frac{11!}{2!}$

Calculating, $11! = 39916800$ $2! = 2$

Therefore: $\text{Arrangements} = \frac{39916800}{2} = 19958400$

So, there are 19,958,400 different arrangements.

Using the factor theorem, since $x + 1$ is a factor, we know: $P(-1) = (-1)^3 + a(-1)^2 + b(-1) - 12 = 0$

This simplifies to: $-1 + a - b - 12 = 0\Rightarrow a - b = 13 \quad (1)$

Now using the remainder theorem: $P(2) = 2^3 + a(2^2) + b(2) - 12 = -18$

This gives: $8 + 4a + 2b - 12 = -18\Rightarrow 4a + 2b - 4 = -18\Rightarrow 4a + 2b = -14\Rightarrow 2a + b = -7 \quad (2)$

Now, we can solve equations (1) and (2): Substituting (1) into (2):

• From (1): $b = a - 13$ [-15][= 4\rightarrow \text{substituted into equation (2)}]
• Thus, $2a + (a - 13) = -7$ $3a - 13 = -7$ $3a = 6\Rightarrow a = 2$ $b = 2 - 13 = -11$

Thus, $a = 2$ and $b = -11$.